Rearranging Equations Watch

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I'm just going over rearranging equations to make another term the subject and just need a little help on understanding the order of operations for rearranging them. I know you generally work backwords from BIDMAS but if you have fractions or a fraction like 1/2 multiplying other terms do you get rid of them first?

Edit: Also another enquiry. If I had for example A=1/2(a+b)h
and I then did A/h=1/2(a+b) would I then do 2A/h= a+b OR 2A/2h= a+b?
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icapturethecastle
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no, you don't neccessarily have to get rid of them
do u have a specific example?
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Unknown?
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Well say for example s= u+v (all over 2) and then multiply it by t.

I got rid of the two first so I had 2s= u + vt (though I'm not sure if the t now just multiplys the v as before I would have added u+v then divided them by 2 and then multiply them. Anyway my final answer is 2s-vt= u since I was trying to make u the subject.

Another one I'm wondering about 1/R= 1/R1 +1/R2 (the 1 and 2 after the R's are small at the bottom of the letter like in a chemistry equation). I inversed them so I had R= R1 + R2 but the book and teacher did something else I can't remember and apparently my solution is wrong.
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Small123
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(Original post by Unknown?)
Well say for example s= u+v (all over 2) and then multiply it by t.

I got rid of the two first so I had 2s= u + vt (though I'm not sure if the t now just multiplys the t as before I would have added u+v then divided them by 2 and then multiply them. Anyway my final answer is 2s-vt= u since I was trying to make u the subject.

Another one I'm wondering about 1/R= 1/R1 +1/R2 (the 1 and 2 after the R's are small at the bottom of the letter like in a chemistry equation). I inversed them so I had R= R1 + R2 but the book and teacher did something else I can't remember and apparently my solution is wrong.
For the second question you cannot take the inverses in the manner that you have because if you could then \dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} \Rightarrow 2=4+4 which isn't true. Generalebriety explains it better. You could times by RR_1 R_2 and you should be able to take it from there.
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generalebriety
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(Original post by Unknown?)
Well say for example s= u+v (all over 2) and then multiply it by t.

I got rid of the two first so I had 2s= u + vt (though I'm not sure if the t now just multiplys the t as before I would have added u+v then divided them by 2 and then multiply them.
You've spotted your own mistake here. t multiplied all of (u+v)/2. Can you correct it?

(Original post by Unknown?)
Another one I'm wondering about 1/R= 1/R1 +1/R2 (the 1 and 2 after the R's are small at the bottom of the letter like in a chemistry equation). I inversed them so I had R= R1 + R2
When you're doing things like this, go very slowly, take each step at a snail's pace, and think very precisely about what you're doing. What you tried to do was take the reciprocal of both sides, which is fine, but ends up looking like this:

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

\left(\frac{1}{R}\right)^{-1} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1}

R = \dfrac{1}{\frac{1}{R_1} + \frac{1}{R_2}}

Unfortunately for you, the right hand side isn't equal to R_1 + R_2 - there's no reason it should be (try a numerical example - put R_1 = R_2 = 1 or something). Instead of doing all this, try writing 1/R_1 + 1/R_2 as one fraction over a common denominator right at the start.
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generalebriety
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(Original post by Small123)
For the second question you cannot take the inverses as this goes against fractional laws.
If you're going to say things like this, at least explain why they're true - otherwise people will get the impression that maths is just a bunch of rules to be learnt, and it's not.
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Small123
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(Original post by generalebriety)
If you're going to say things like this, at least explain why they're true - otherwise people will get the impression that maths is just a bunch of rules to be learnt, and it's not.
I understand that I was unclear after reading your post. You can inverse just not in the sense that the OP has done.
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generalebriety
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(Original post by Small123)
I understand that I was unclear after reading your post. You can inverse just not in the sense that the OP has done.
Yeah, sure, just please don't talk about "fractional laws" as if they're something mysterious we know about and the OP doesn't.
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Unknown?
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(Original post by generalebriety)
You've spotted your own mistake here. t multiplied all of (u+v)/2. Can you correct it?


When you're doing things like this, go very slowly, take each step at a snail's pace, and think very precisely about what you're doing. What you tried to do was take the reciprocal of both sides, which is fine, but ends up looking like this:

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

\left(\frac{1}{R}\right)^{-1} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1}

R = \dfrac{1}{\frac{1}{R_1} + \frac{1}{R_2}}

Unfortunately for you, the right hand side isn't equal to R_1 + R_2 - there's no reason it should be (try a numerical example - put R_1 = R_2 = 1 or something). Instead of doing all this, try writing 1/R_1 + 1/R_2 as one fraction over a common denominator right at the start.
Thanks so would the answer to the first question be 2s/t then -v=u. Also how would I write a number being divided by another fraction as I don't think I can do that in an exam/test.
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Small123
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(Original post by Unknown?)
Thanks so would the answer to the first question be 2s/t then -v=u. Also how would I write a number being divided by another fraction as I don't think I can do that in an exam/test.
Times both sides by R_1 R_2 to get to a simplified fraction.
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generalebriety
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(Original post by Unknown?)
Thanks so would the answer to the first question be 2s/t then -v=u. Also how would I write a number being divided by another fraction as I don't think I can do that in an exam/test.
I told you how to simplify that at the end.
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Unknown?
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Okay I've got another question. :o: If I've got to rearrange an equation and all of one side of the equation has been square rooted would I get rid of the square root first? I normally try imagining the correct way in my head like if 64 square rooted = 8 then if I cancel out out the square root then 64=8 squared.
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generalebriety
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(Original post by Unknown?)
Okay I've got another question. :o: If I've got to rearrange an equation and all of one side of the equation has been square rooted would I get rid of the square root first? I normally try imagining the correct way in my head like if 64 square rooted = 8 then if I cancel out out the square root then 64=8 squared.
Yes. Try to work "from the outside in". If the square root is covering everything, then get rid of that first. (Remember also that taking a square root is equivalent to raising to the power of 1/2, so this is really a question of brackets and indices, which are catered for in BIDMAS.)
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